L'Hôpital's rule (0/0 form, one-sided): Let ff and gg be continuous on [a,b)[a,b) and differentiable on (a,b)(a,b), and assume that g′(x)≠0g'(x)\neq 0 for all x∈(a,b)x\in(a,b). Suppose

lim⁡x→a+f(x)=0andlim⁡x→a+g(x)=0,\lim_{x\to a^+} f(x)=0 \quad\text{and}\quad \lim_{x\to a^+} g(x)=0,

and that g(x)≠0g(x)\neq 0 for all xx sufficiently close to aa with x>ax>a. If the limit

L=lim⁡x→a+f′(x)g′(x)L=\lim_{x\to a^+}\frac{f'(x)}{g'(x)}

exists (as a finite number or as ±∞\pm\infty), then the limit

lim⁡x→a+f(x)g(x)\lim_{x\to a^+}\frac{f(x)}{g(x)}

also exists and equals LL.

Remarks

Analogous statements hold for left-hand limits and for the ∞/∞\infty/\infty indeterminate form, and there are versions for . The proof is based on the and is formulated in terms of .